\(\int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [156]

Optimal result

Integrand size = 23, antiderivative size = 33 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 3852, 8} \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[(b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(7/2),x]

[Out]

(b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{\sqrt {\cos (c+d x)}} \\ & = -\frac {\left (b \sqrt {b \cos (c+d x)}\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d \sqrt {\cos (c+d x)}} \\ & = \frac {b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(7/2),x]

[Out]

((b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {b \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(30\)
risch	\(\frac {2 i b \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(39\)

[In]

int((cos(d*x+c)*b)^(3/2)/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

b*sin(d*x+c)*(cos(d*x+c)*b)^(1/2)/d/cos(d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {b \cos \left (d x + c\right )} b \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{\frac {3}{2}}} \]

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

sqrt(b*cos(d*x + c))*b*sin(d*x + c)/(d*cos(d*x + c)^(3/2))

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(3/2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, b^{\frac {3}{2}} \sin \left (2 \, d x + 2 \, c\right )}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \]

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

2*b^(3/2)*sin(2*d*x + 2*c)/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*d)

Giac [F]

\[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(3/2)/cos(d*x + c)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 13.91 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+1{}\mathrm {i}\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int((b*cos(c + d*x))^(3/2)/cos(c + d*x)^(7/2),x)

[Out]

(b*(b*cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 1i))/(d*cos(c + d*x)^(1/2)*(cos(2*c + 2*d*
x) + 1))